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AI-GEOSTATS: Calculating averages

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  • Allison, Craig
    Hello, I was hoping advice could be given on the following problem. We store nickel (eg 0.40 Ni %) and mill recovery (55.45%) in our mine model. From this we
    Message 1 of 3 , Jan 5, 2003
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      Hello,

      I was hoping advice could be given on the following problem. We store nickel
      (eg 0.40 Ni %) and mill recovery (55.45%) in our mine model. From this we
      calculate a mill recovered nickel (eg 0.22 RecNi %) and store this in the
      model. All 3 model items are stored to two decimal places. The rounding is
      to three decimal places (i.e. +/-0.005).

      We create ore-cuts based on the tonnes-weighted average of the Mill
      Recovered Nickel (using an Ore-material classification scheme based off Mill
      Recovered Nickel).

      My problem is given below and highlighted in green. I have assumed tonnage
      weighting in each row is 1.


      Nickel Mill Recovered
      Head Recovery Nickel
      Grade % (Ni %)
      0.53 60 0.3180
      0.65 62 0.4030
      0.32 20 0.0640
      0.8 70 0.5600
      0.56 63 0.3528
      0.45 55 0.2475
      0.58 68 0.3944
      Average 0.555714286 56.85714286 0.3342

      Average Nickel Head Grade
      divided by Average Mill Recovery ====> 0.315963265



      It would appear that the averaging of the recovered nickel (0.3342) gives a
      different result to dividing the average Nickel Head Grade by the Average
      Mill Recovery (0.315963265). These averages are generated in Microsoft Excel
      so it doesn't seem to be a rounding error. We believe the first average is
      wrong and it would appear that the greater the variability of the data set,
      the greater the difference between the two calculation averages. If
      possible, would anyone know what is happening here and which is the correct
      approach.

      Thank-you in appreciation.





      Craig Allison

      MKO Resource Evaluation Geologist




      [Non-text portions of this message have been removed]
    • Digby Millikan
      Calculating averagesCraig, Don t you mean multiply Average Nickel Head grade by Average Mill recovery. Your first figure 0.3342 is correct but your figure
      Message 2 of 3 , Jan 5, 2003
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        Calculating averagesCraig,
        Don't you mean multiply Average Nickel Head grade by Average Mill
        recovery. Your first figure 0.3342 is correct but your figure 0.315963265
        is incorrect. It is your assumption that your average mill recovery is
        tonneage recovery, i.e. you averaged the mill recovery as if it where
        recovery per tonne of ore instead of per % of metal so your calculation
        of 0.5557*0.56857 is incorrect.

        Your average mill recovery should be calculated

        (0.53*60)+(0.65*62)+(0.32*20)+(0.8*70)+(0.56*63)+(0.45*55)+(0.58*0.68)/
        (0.53+0.65+0.32+0.8+0.56+0.45+0.58) =0.601465

        So your original table should look like;

        Nickel Mill Recovered
        Head Recovery Nickel
        Grade % (Ni %)
        0.53 60 0.3180
        0.65 62 0.4030
        0.32 20 0.0640
        0.8 70 0.5600
        0.56 63 0.3528
        0.45 55 0.2475
        0.58 68 0.3944
        Average 0.555714286 60.14652956 0.3342

        Average Nickel Head Grade
        divided by Average Mill Recovery ====> 0.334242857


        Regards Digby Millikan
        Geolite Mining Systems
        digbym@...
        http://www.users.on.net/digbym


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      • Isobel Clark
        Hi Craig The average of a product will only equal the product of the averages if the two numbers are completely independent of one another. This is exactly
        Message 3 of 3 , Jan 5, 2003
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          Hi Craig

          The average of a product will only equal the product
          of the averages if the two numbers are completely
          independent of one another.

          This is exactly analogous to the calculation of a
          covariance in statistics.

          Isobel
          http://uk.geocities.com/drisobelclark

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